From Game Rules to a Distribution
In Interpreting a Probability Distribution in Context, you learned to explain what a probability says about a random variable in a chance process. To make that kind of interpretation, you first need a distribution. A game description often gives the ingredients in words: how much it costs to play, what prizes are possible, and how likely each outcome is. The task is to turn those details into possible values of a random variable and their probabilities.
For games, it is useful to define the random variable \(X\) as the player's net winnings on one play. Net winnings account for both the prize received and the cost paid. This matters because a prize is not necessarily a gain: a player might win a small prize but still spend more to play than the prize is worth.
A game description may give outcomes as tickets, die faces, spinner sectors, or another chance mechanism. Use those outcomes to find how likely each result is. Do not assume that different prize amounts are equally likely just because each prize is listed once. The underlying chance outcomes determine the probabilities.
A Reliable Translation Process
Work in two passes. First identify the chance outcomes and the prize attached to each. Then convert each prize to net winnings by subtracting the cost. If several chance outcomes produce the same net winnings, combine their probabilities into one row of the distribution. This is how you move from the details of the game to a table for \(X\).
State that \(X\) is the player's net winnings on one play, or give another precise meaning if the question specifies a different quantity.
Record the tickets, faces, sectors, or other outcomes and the prize each one awards.
For each prize, calculate \(X=\text{prize value}-\text{cost to play}\). Keep a loss negative.
Use the chance mechanism to determine the probability of each net result. Add probabilities when different outcomes lead to the same value of \(X\).
Include every possible net result, confirm each probability is between 0 and 1, and check that all probabilities add to 1.
For a set of equally likely outcomes, a result's probability is the number of outcomes producing it divided by the total number of outcomes. The denominator is the total number of equally likely chance outcomes—not the number of different prize amounts. This counting approach is the same basic distribution-building idea used in Building a Distribution from a Sample Space; here, the verbal rules tell you how to assign the random variable's value.
Worked Example: A Prize Ticket Game
An invented ticket game costs $3 to play. There are 20 equally likely tickets: 10 award no prize, 5 award a $2 prize, 3 award a $5 prize, and 2 award a $20 prize. Let \(X\) be the player's net winnings on one ticket.
State. Build the probability distribution of \(X\).
Plan. The ticket outcomes are equally likely, so find each probability by dividing the number of tickets in that prize category by 20. For each category, subtract the $3 cost from the prize value.
Do. The net results are $0 - $3 = -$3, $2 - $3 = -$1, $5 - $3 = $2, and $20 - $3 = $17. The corresponding probabilities are calculated as follows:
The distribution is:
| Net winnings, \(x\) | \(P(X=x)\) |
|---|---|
| -$3 | 0.50 |
| -$1 | 0.25 |
| $2 | 0.15 |
| $17 | 0.10 |
Check the probabilities: \(0.50+0.25+0.15+0.10=1.00\). Each probability is between 0 and 1, and the table includes all four prize categories, so it accounts for all 20 tickets.
Conclude. \(X\) can be -$3, -$1, $2, or $17. For example, \(P(X=-$1)=0.25\) means that one quarter of the equally likely tickets award a $2 prize, leaving the player with a net loss of $1 after paying to play.
Watch the Chance Outcomes, Not Just the Prize List
The prize list alone does not tell you the distribution. A prize awarded on many spinner sectors is more likely than one awarded on only one sector, even if the game description lists each prize just once. Similarly, different chance outcomes can lead to the same net result. In that case, the probabilities for those outcomes belong together in one distribution row.
The entry cost affects the values of \(X\), but it does not change the probabilities of the chance outcomes. If the game costs $1, every prize value shifts down by $1 when expressed as net winnings. The chance of receiving each prize remains determined by the die, spinner, or other mechanism.
Worked Example: A Six-Sided Prize Die
An invented game costs $1 to play. A fair six-sided die awards no prize on a 1 or 2, a $1 prize on a 3 or 4, a $3 prize on a 5, and a $7 prize on a 6. Let \(X\) be the player's net winnings on one roll.
State. Find the probability distribution of \(X\).
Plan. Because the die is fair, each of its six faces is equally likely. Translate each prize into net winnings by subtracting $1. Count how many faces lead to each net result, including faces that share a prize category.
Do. A $0 prize gives net winnings of \($0-$1=-$1\), and that result occurs on faces 1 and 2. A $1 prize gives \($1-$1=$0\), occurring on faces 3 and 4. A $3 prize gives \($3-$1=$2\), occurring on face 5. A $7 prize gives \($7-$1=$6\), occurring on face 6. Therefore:
The distribution is:
| Net winnings, \(x\) | \(P(X=x)\) |
|---|---|
| -$1 | \(\frac{1}{3}\) |
| $0 | \(\frac{1}{3}\) |
| $2 | \(\frac{1}{6}\) |
| $6 | \(\frac{1}{6}\) |
The probabilities total \(\frac{1}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{6}=1\). Each of the six die faces appears in exactly one prize category, so no face has been omitted or counted twice.
Conclude. On one roll, the player's net winnings are -$1, $0, $2, or $6, with the probabilities shown. In particular, \(X=0\) means breaking even: the player receives a $1 prize but has also paid $1 to play. It does not mean that the player received no prize.
Different Outcomes Can Lead to the Same Net Result
A chance outcome and a value of \(X\) are not always the same thing. Several outcomes may lead to one prize, and therefore to one net result. For instance, on a spinner, multiple sectors may award the same amount. The probability for that net result is the probability that the spinner lands on any of those sectors.
This is also why a distribution table should have one row for each possible value of \(X\), not necessarily one row for each raw outcome in the game. If two distinct outcomes both result in net winnings of -$2, they contribute to the same event \(X=-$2\). Add their probabilities, or count both outcomes together when the basic outcomes are equally likely.
Worked Example: A Spinner with Unequal Prize Counts
An invented spinner has eight equal-sized sectors. Three sectors award no prize, two award $2, two award $4, and one awards $10. It costs $3 to spin. Let \(X\) be the player's net winnings from one spin.
State. Construct the distribution of \(X\).
Plan. Equal-sized sectors make the eight sectors equally likely. For each prize, subtract the $3 cost. Then use the number of sectors for that prize over eight to find the probability of the resulting net value.
Do. The net values are -$3, -$1, $1, and $7. Their probabilities are:
The distribution is:
| Net winnings, \(x\) | \(P(X=x)\) |
|---|---|
| -$3 | 0.375 |
| -$1 | 0.250 |
| $1 | 0.250 |
| $7 | 0.125 |
As a check, \(0.375+0.250+0.250+0.125=1.000\). The three no-prize sectors are all included in \(P(X=-3)\), and the remaining five sectors are included in the other three rows.
Conclude. The player's net winnings are -$3, -$1, $1, or $7. The probability of a net loss of $3 is \(3/8\), because three of the eight equal sectors award no prize. Although four different prizes or prize categories are described, the chance of each depends on the number of sectors assigned to it.
Common Mistakes and AP Exam Tips
- Listing the prize instead of net winnings. If the question asks for the player's net result, a $2 prize in a game that costs $3 is not a gain of $2. The net winnings are \($2-$3=-$1\).
- Making every listed prize equally likely. A list of prize amounts does not establish that each prize has the same probability. Use the number of tickets, faces, or sectors assigned to each prize.
- Ignoring a no-prize outcome. “No prize” still has a net result. If the game has a cost, no prize produces a loss equal to that cost.
- Confusing no prize with breaking even. Breaking even means the prize exactly covers the cost. A no-prize result is usually a loss when playing costs money.
- Leaving repeated results in separate rows. If different outcomes produce the same value of \(X\), combine their probabilities so the table gives \(P(X=x)\) for each distinct value \(x\).
- Using the number of prize categories as the denominator. For equally likely chance outcomes, divide by the total number of tickets, faces, or sectors—not by the number of different prizes.
- Failing to check completeness. A table can have probabilities that add to 1 and still mislabel a net result. Check both the arithmetic and the meaning of every row against the game rules.
A clear AP response defines \(X\), shows how each prize becomes a net result, and explains how the chance mechanism determines each probability. When checking the table, state that every possible result is included, every probability is between 0 and 1, and the probabilities sum to 1. That makes the reasoning visible rather than leaving the reader to guess how a row was formed.
Key Takeaway
A game's verbal rules can be turned into a probability distribution by linking chance outcomes to prizes, subtracting the cost to play, and assigning each net result its probability. Keep losses negative, use the chance mechanism to determine probabilities, and combine outcomes that produce the same net winnings.
Check Your Understanding
For each game, focus on the chance outcomes, the entry cost, and the player's net winnings.
- A game costs $2. Of 12 equally likely tickets, 6 award no prize, 4 award $2, and 2 award $8. Define \(X\) as net winnings and construct its probability distribution.
- A fair die game costs $3. A roll of 1 or 2 awards $1, a roll of 3, 4, or 5 awards $3, and a roll of 6 awards $9. What are the possible net winnings and their probabilities?
- A spinner has 10 equal sectors: 4 award $0, 3 award $1, 2 award $5, and 1 awards $10. If it costs $2 to spin, what is \(P(X=-$2)\)? Explain how you found it.
- In a game that costs $4, a player receives a $4 prize. What are the player's net winnings, and what does that result mean in context?
- Why should two different chance outcomes that both produce net winnings of $1 appear together in one row of a distribution table?